I have a problem with some consideration that Mumford does about very ample line bundles in the prove of Riemann-Roch theorem.
Namely, he says that if we consider a very ample line bundle $L=O(D)$ on $A$ Abelian variety, we get $\sigma_0,...,\sigma_g$ ($g=dimA$) sections such that
i) they have no common zeroes (and here I suppose he's using the fact that $L$ is globally generated and open immersions separate points)
ii) the divisors of zero of $\sigma_1,...,\sigma_g$ intersect transversally at $(D^g)$ points (and this is also clear as we are dealing with a variety over an algebraically closed field)
condition i) implies that there is a morphism $\phi: A \to \mathbb{P}^g$ defined by $(\sigma_0(x),...,\sigma_g(x))$
But after that he uses that L is the pullback of $O(1)$ by $\phi$, what is not clear to me is why should be straightforward that $\phi$ is exactly the morphism defined by ampleness.
Could anyone help me? Thanks!
These type of questions should be dealt with using standard universal properties. Let $X$ be any variety, $L$ a line bundle and assume $s_0,\ldots, s_g\in H^0(L)$. Then we get a homomorphism $\mathcal{O}_X^{g+1}\to L$, using these sections. These sections have no common zeros is equivalent to saying the above map is onto. On the other hand, giving such a surjection is equivalent to giving a morphism $f:X\to\mathbb{P}^g$ such that the pull back under $f$ of the surjection $\mathcal{O}_{\mathbb{P}^g}^{g+1}\to\mathcal{O}_{\mathbb{P}^g}(1)$ is the map we started with. If in addition $L$ is ample with $X$ projective, we see that $f$ is a finite map onto its image. In your case, since $\dim X=g$, we see that the map is finite onto all of the projective space.